The algebraic curves of planar polynomial differential systems with homogeneous nonlinearities
Vladimir M. Cheresiz
1and Evgenii P. Volokitin
B1, 21Sobolev Institute of Mathematics, Novosibirsk, 630090, Russia
2Novosibirsk State University, Novosibirsk, 630090, Russia
Received 7 March 2021, appeared 15 July 2021 Communicated by Armengol Gasull
Abstract. We consider planar polynomial systems of ordinary differential equations of the form ˙x=x+Pn(x,y), ˙y=y+Qn(x,y), wherePn(x,y), Qn(x,y) are homogeneous polynomials of degreen. We study the algebraic and non-algebraic invariant curves of these systems with emphasis on limit cycles.
Keywords: polynomial systems, algebraic limit cycles.
2020 Mathematics Subject Classification: 34C05, 34A34.
1 Introduction
The problem of studying the limit cycles is one of the central problems in the theory of ordi- nary differential equations. A significant subarea in this area is the study of the limit cycles of autonomous planar polynomial systems
˙
x= P(x,y), y˙= Q(x,y). (1.1) Here P(x,y), Q(x,y) are real polynomials of the variables x, y; t ∈ R acts as an inde- pendent variable. The degree of the system is the maximum of the degrees of the polynomi- alsP(x,y), Q(x,y).
A limit cycle of system (1.1) is a periodic solution whose trajectory is isolated among the trajectories of all periodic solutions. A limit cycle of system (1.1) is called algebraic of de- greemif it is the real oval of an irreducible algebraic curveH(x,y) =0 of degreem.
The problems of finding algebraic solutions to polynomial systems, in particular, of al- gebraic cycles, goes back to H. Poincaré and J.-G. Darboux, and are actively developing at present (see [5] and the literature cited therein).
In the present article, we study these problems in application to a differential system of the form
˙
x =x+Pn(x,y), y˙ =y+Qn(x,y), (1.2)
BCorresponding author. Email: volok@math.nsc.ru
which we call a Darboux-type system. HerePn(x,y), Qn(x,y)are homogeneous real polyno- mials of degreen of the variablesx, y. Systems of such a kind appeared in Darboux’s works on geometry.
We study necessary and sufficient conditions for the existence of a hyperbolic limit cycle for system (1.2), and this cycle turns out to be unique. The remaining trajectories (except for the singular point at the origin) have a limit cycle as theα- orω-limit set and cannot be algebraic curves.
We prove that the degree of an algebraic limit cycle of system (1.2) is equal to 2 and obtain necessary and sufficient conditions for the existence of an algebraic limit cycle for (1.2).
The obtained results are illustrated by examples.
2 The main part
Consider a Darboux-type system of a more general kind than (1.2):
˙
x= sx+Pn(x,y), y˙ = sy+Qn(x,y), s6=0, n>1. (2.1) Consider the functions
f(ϑ) =cosϑPn(cosϑ, sinϑ) +sinϑQn(cosϑ, sinϑ), g(ϑ) =cosϑQn(cosϑ, sinϑ)−sinϑPn(cosϑ, sinϑ). Theorem 2.1.
(1) If n is even then system(2.1)has no periodic solutions.
(2) If n is odd and g(ϑ)has zeros on[0, 2π]then system(2.1)has no periodic solutions.
(3) If there is a closed trajectory of system(2.1), then it contains inside itself the only singular point that coincides with the origin.
(4) System(2.1)has at most one limit cycle.
(5) For system (2.1) to have a unique limit cycle Γ, it is necessary and sufficient that the following conditions hold:
g(ϑ)6=0, ϑ∈[0, 2π]; sg(0)
Z 2π
0
f(ϑ)
g(ϑ)dϑ<0. (2.2) (6) The cycleΓis hyperbolic.
(7) If the cycle Γ is algebraic then its degree is equal to 2 and it is defined by an algebraic curve H(x,y) =0, H(x,y) =1+ax2+2bxy+cy2.
Proof. Items (1)–(5) are proved in [1,7].
We will briefly give fragments of this proof in order to use them for proving items (6)–(7), which supplement the results of [1,7] concerning Darboux-type systems.
Henceforth, unless otherwise specified, we assume that n is odd and conditions (4) are fulfilled.
After passing to the polar coordinatesx=rcosϑ,y=rsinϑ, system (2.1) turns into the sys- tem
r˙= sr+rnf(ϑ), ϑ˙ =rn−1g(ϑ), (2.3) which we replace by the linear equation
dρ
dϑ = (n−1)f(ϑ)
g(ϑ) ρ+s(n−1)
g(ϑ) , ρ=rn−1, (2.4) where the functions f(ϑ), g(ϑ)are defined above.
The periodic solutions to this equation (with period T = 2π) generate periodic solutions to system (2.1).
Introduce the function
F(ϑ) =exp
(n−1)
Z ϑ
0
f(τ) g(τ)dτ
. Denote the solution to equation (2.4) with ρ(0) =ρ0byρ(ϑ;ρ0):
ρ(ϑ;ρ0) =
ρ0+s(n−1)
Z ϑ
0
dτ g(τ)F(τ)
F(ϑ). (2.5)
For a periodic solution, we must takeρ0=ρ∗0 from the condition ρ(2π;ρ0)=ρ0. In the case under consideration, there exists a unique such value:
ρ∗0 =s(n−1) F(2π) 1−F(2π)
Z 2π
0
dτ
g(τ)F(τ). (2.6)
For a solutionρ = ρ(ϑ,ρ0∗)to define a periodic solution to (2.1), the condition ρ(ϑ,ρ∗0) >
0, ϑ∈ [0, 2π]must be fulfilled. This condition holds by (2.2).
The orbits of system (2.1) have the parametric definition x = n−1
q
ρ(ϑ;ρ0)cosϑ, y= n−1 q
ρ(ϑ;ρ0)sinϑ, ϑ∈[0, 2π], (2.7) whereρ(ϑ;ρ0)is from (7).
For a periodic orbit, we must takeρ0=ρ∗0 in (2.6).
The cycle under consideration is hyperbolic. For showing this, calculate the derivative of the solution (7) with respect toρ0 forϑ=2π at the pointρ0 =ρ∗0.
µ= ∂ρ(ϑ;ρ0)
∂ρ0
{ϑ=2π,ρ0=ρ∗0}
= F(2π). By (2.2), we have thatµ6=1, i.e., the cycleΓis hyperbolic.
Item (6) is proved.
For proving item (7), make use of another trick, proposed in [4].
The direction field of system (2.1) is symmetric with respect to the origin. In this case, the trajectories of this system and the formulas defining them must also possess the symme- try property. In particular, closed algebraic curves are defined by polynomials of the form H(x,y) = h0+h2(x,y) +h4(x,y) +. . . , where h0 = const 6= 0, h2(x,y), h4(x,y), . . . are ho- mogeneous polynomials of even degrees. Without loss of generality, we may assume that h0=1.
If the limit cycle of system (2.1) is algebraic then consider an irreducible polynomial H(x,y) =1+h2(x,y) +· · ·+h2k(x,y)such thatH(x,y) =0 contains the oval defined by (2.7)
withρ0=ρ∗0. H(x,y) =0 is an invariant algebraic curve of (2.1), [4]. ThenH(rcosϑ,rsinϑ) = 0 is an invariant curve of system (2.3). After inserting R= r2 in H(rcosϑ,rsinϑ), we obtain a polynomial ˜H(R,ϑ)of the variableRwhose coefficients arehi(cosϑ, sinϑ).
Note that the polynomial ˜H(R,ϑ) has only positive roots. Moreover, R = (ρ(ϑ;ρ∗0))n−21 is its R-root. Each R-root of ˜H(R,ϑ) generates a solution ρ(ϑ;ρ0) = Rn−21 to (2.4). If ρ0 6=
ρ∗0 then the solution ρ(ϑ;ρ0) takes infinitely many different values at the points ϑ = 2πk for all integersk. On the other hand, ˜H(R, 2πk) = H˜(R, 0)for all integers k; therefore, these polynomials have the same roots. Hence, the polynomial ˜H(R, 0) has infinitely many roots R = (ρ(2πk;ρ0))n−21, which is impossible. We see that the R-root necessarily corresponds to a unique limit cycle system R(n−1)/2 = ρ(ϑ;ρ∗0). Since the polynomial ˜H has only positive roots, we conclude that ˜H(R,ϑ) has one and only one R-root and, thus, it takes the form H˜(R,ϑ) =1+h2(cosϑ, sinϑ)R. ThenH(rcosϑ,rsinϑ) =1+r2h2(cosϑ, sinϑ), which implies that H(x,y) =1+ax2+2bxy+cy2.
Item (7) is proved.
Theorem 2.2. System(2.1)has an algebraic limit cycle H(x,y)≡ 1+h2(x,y) = 0, h2(x,y)< 0if and only if the conditions
Pn
∂h2
∂x +Qn
∂h2
∂y =2s(−h2)n+21; xPn−yQn 6=0 for(x,y)6= (0, 0) are satisfied.
Proof. Recall that if the trajectory of a planar polynomial system of differential equations
˙
x =P(x,y), y˙ =Q(x,y) (1.1) is a part of an irreducible algebraic curve H(x,y) = 0 then there exists a polynomial k(x,y) (cofactor) such that
∂H(x,y)
∂x P(x,y) + ∂H(x,y)
∂y Q(x,y) =k(x,y)H(x,y). (2.8) Obviously, the degree of a cofactor is at most n−1 if n is the maximum of the degrees of the polynomialsP(x,y), Q(x,y).
As follows from Theorem 2.1, for system (2.1), the closed algebraic curve (an algebraic limit cycle) is defined by the polynomial of the form H(x,y) =1+h2(x,y), whereh2(x,y)is a homogeneous polynomial of degree 2.
Condition (2.8) takes the form (sx+Pn) ∂
∂x(1+h2) + (sy+Qn) ∂
∂y(1+h2) =k(1+h2). Settingn=2m+1, we havek =k2+k4+· · ·+k2m, and
(sx+P2m+1)∂h2
∂x + (sy+Q2m+1)∂h2
∂y
=k2+ (h2k2+k4) + (h2k4+k6) +· · ·+ (h2k2m−2+k2m) +h2k2m.
We get
sx∂h2
∂x +sy∂h2
∂y =k2, h2k2+k4 =0, h2k4+k6 =0,
... h2k2m−2+k2m =0, P2m+1
∂h2
∂x +Q2m+1
∂h2
∂y =h2k2m.
Using Euler’s formula for the homogeneous polynomialh2(x,y), we have 2sh2=k2.
Further, we put
k4=−h2k2=−2sh22, k6=−h2k4=2sh32, . . . , k2m=−h2k2m−2=(−1)m+12sh2m. Based on these equalities, we have
P2m+1
∂h2
∂x +Q2m+1
∂h2
∂y =2s(−1)m+1h2m+1. The last equality proves the invariance of the curveH(x,y) =0.
There are no singular points on the curve.
The theorem is proved.
In the case of a cubic Darboux system (n = 3), we can give an exhaustive solution the problem under consideration.
In [2], Theorem 3.2 was proved, which classifies planar homogeneous cubic vector fields.
In our case, this theorem has the following consequence:
Proposition 2.3. The system
˙
x =x+P3(x,y), y˙ =y+Q3(x,y) (2.9) has a limit cycle only if there exists a linear transformation σ ∈ GL(2;R) and a time scaling taking system(2.9)into the system of the form
˙
x=sx+p1x3+ (p2−α)x2y+p3xy2−αy3≡ sx+P˜3(x,y),
˙
y=sy+αx3+p1x2y+ (p2+α)xy2+p3y3≡sy+Q˜3(x,y), p1, p2, p3, s∈R, α=±1.
(2.10)
For system (2.10),
f(ϑ) = 1
2(p1+p3+ (p1−p3)cos 2ϑ+p2sin 2ϑ), g(ϑ)≡α, sg(0)
Z 2π
0
f(ϑ)
g(ϑ)dϑ=πs(p1+p3).
By Theorem 2.1, we infer that, for the existence of a unique hyperbolic limit cycle for sys- tem (2.10), it is necessary and sufficient that
s(p1+p3)<0. (2.11)
Consider the question of the existence of an algebraic limit cycle for system (2.10).
Suppose that system (2.10) has a quadratic limit cycle H=1+h2≡1+ax2+2bxy+cy2. Condition (2.8) takes the form
∂H
∂x (sx+P˜3) + ∂H
∂y(sy+Q˜3) =k2H.
From Theorem2.2 we obtain
P˜3∂h2
∂x +Q˜3∂h2
∂y =2sh22. (2.12)
Equating the coefficients at the same degrees of the variables x,y on the left- and right- hand sides of (2.12), after easy transformations, we obtain the system of equalities
−a2s+ap1+αb=0,
−4abs+ap2−αa+2bp1+αc=0,
−2acs+ap3−4b2s+2bp2+cp1 =0,
−αa−4bcs+2bp3+cp2+αc=0,
−αb−c2s+cp3 =0.
(2.13)
Here and below, we have used the Mathematica system for implementing symbols and numerical calculations.
System (2.13) can be regarded as an inhomogeneous system of linear equations AX = B with respect to the variablesp1,p2,p3with the parametersa,b,c,s,α
A=
a 0 0
2b a 0
c 2b a
0 c 2b
0 0 c
, B=
a2s−αb α(a−c) +4abs
2acs+4b2s α(a−c) +4bcs
αb+c2s
, X=
p1 p2
p3
.
The system is solvable if and only if the rank of the matrix Ais equal to the rank of the ex- tended matrix(A|B).
Obviously, a 6= 0 in (2.13). Therefore, rank A = 3. Then all minors of order 4 in (A|B) must be zero. These minors are
(A|B)1 =α(a2c2−8ab2c−2ac3+8b4+4b2c2+c4), (A|B)2 =α(−3a2bc+4ab3+2abc2+bc3),
(A|B)3 =α(−a3c+2a2b2+2a2c2−ac3+2b2c2), (A|B)4 =α(a3b+2a2bc−3abc2+4b3c),
(A|B)5 =α(a4−2a3c+4a2b2+a2c2−8ab2c+8b4),
where (A|B)i stands for the minor obtained from (A|B) by deleting the ith row. Note that the obtained expressions do not depend onsand containαas a factor.
Sinceα=±1, it suffices to consider the system of homogeneous equations a2c2−8ab2c−2ac3+8b4+4b2c2+c4=0,
−3a2bc+4ab3+2abc2+bc3=0,
−a3c+2a2b2+2a2c2−ac3+2b2c2=0, a3b+2a2bc−3abc2+4b3c=0, a4−2a3c+4a2b2+a2c2−8ab2c+8b4=0.
(2.14)
The second and fourth equations have the form
b(−3a2c+4ab2+2ac2+c3) =0, b(a3+2a2c−3ac2+4b2c) =0. (2.15) Case 1. b=0.
In this case, system (2.14) is reduced to the system
c2(a−c)2= ac(a−c)2 =a2(a−c)2 =0,
which implies that we have a nonzero solutiona= c6=0, b=0 to system (2.14).
Case 2.
If b6= 0 then −3a2c+4ab2+2ac2+c3 = a3+2a2c−3ac2+4b2c = 0, which implies that (a−c)(a+c)3= 0. Ifa=cthen the fourth equation of (2.14) givesc=0, and hence the fifth equation yields b = 0, which contradicts the assumptionb 6= 0. If a = −cthen the only real solution to (2.15) isa =b= c=0, which again leads to a contradiction.
For the above-found values of the parametersa, b, c, system (2.13) is reduced to the form ap1=a2s, ap2=0, ap1+ap3 =2a2s, ap2 =0, ap3 =a2s, a6=0,
and has a nonzero solution p1 = p3 6=0, p2=0, a= c= p1/s, b=0.
Then H(x,y) = 1+ ps1(x2+y2), and the algebraic curve H(x,y) = 0 defines a real oval (circle) under the condition p1/s<0 (cf. (2.11)).
Thus, we have proved
Theorem 2.4. System(2.10)admits a hyperbolic algebraic cycle if and only if
p1 = p3, p2=0, p1s<0. (2.16) Moreover, the cycle is defined by the algebraic curve
H≡1+ p1
s (x2+y2) =0.
System (2.10) for which conditions (2.16) are fulfilled has the form
˙
x =sx+px3−αx2y+pxy2−αy3, y˙ =sy+αx3+px2y+αxy2+py3. (2.17) Putδ=p−s/p.
A straightforward check shows that
x(t) =δcosδ2t, y(t) =αδsinδ2t
is a periodic solution to system (2.17). This solution is a suitable parametrization for the circle H=0 mentioned in the theorem. The period of the obtained cycle is equal to T=2π/δ2.
The cycle is stable ifs>0 and unstable ifs<0.
Using Therem2.2 and Theorem2.4, in [8], we proved
Theorem 2.5. System(2.10)has an algebraic limit cycle if and only if the coefficients pij, qij, i,j= 0, 1, 2, 3, i+j=3are representable as
p30 =−s c2+d2
(α(ac+bd) +p(ad−bc)) (bc−ad)3 ,
p21 =sa2(α(3c2+d2)+2cdp) +2ab(2αcd+p(d2−c2)) +b2(α(c2+3d2)−2cdp)
(bc−ad)3 ,
p12 =−s(a2+b2)(3αac+adp+3αbd−bcp) (bc−ad)3 , p03 =sα(a2+b2)2
(bc−ad)3, q30= −sα c2+d22
(bc−ad)3 , q21 =s(c2+d2)(3α(ac+bd)−p(ad−bc))
(bc−ad)3 ,
q12 =−sa2(α(3c2+d2)−2cdp) +2ab(2αcd+p(c2−d2)) +b2(α(c2+3d2)+2cdp)
(bc−ad)3 ,
q03 =s(a2+b2)(α(ac+bd)−p(ad−bc)) (bc−ad)3 , where a,b,c,d,p,s ∈R, ad−bc6=0, ps<0, α=±1.
Moreover, the cycle is defined by the algebraic curve H≡1+ ps
(ad−bc)2((ay−cx)2+ (by−dx)2). Consider several examples illustrating the obtained results.
Example 2.6. ([1]).
˙
x=−x+x3−x2y+xy2−y3, y˙ =−y+x3+x2y+xy2+y3. (2.18) The system can be written down in the form (2.10) if we take s = −1, p1 = p3 = 1, p2 = 0, α = 1. Hence, by Theorem 2.4, system (2.18) has the hyperbolic algebraic limit cycle 1−x2−y2 =0. The cycle is unstable since it contains a stable singular point, the origin. This cycle was presented in [1].
Example 2.7.
˙
x=x−2x2y−4xy2−2y3, y˙ =y+2x3+2xy2−4y3. (2.19) We have
f(ϑ) =−4 sin2ϑ, g(ϑ)≡2, F(ϑ) =esin 2ϑ−2ϑ.
By Theorem2.1, we conclude that the system has a hyperbolic stable limit cycle.
By (2.5), (2.6), the cycle is written as r= √
ρ, ρ =
ρ∗0+
Z ϑ
0 e2τ−sin 2τdτ
esin 2ϑ−2ϑ, ρ0∗= 1
e4π−1 Z 2π
0 esin 2τ−2τdτ.
An attempt to find a quadratic limit cycle in the form 1+ax2+2bxy+cy2 = 0 leads to the system of equations
a2−2b=0, a+2ab−c=0, 2a+2b2+ac=0, a+4b−c+2bc=0, 2b+4c+c2=0.
Express b, cin terms of a from the first two equations and insert them in the remaining three equations. After easy calculations, it is not hard to see that the system has only the zero solution a=b=c=0. Hence, the limit cycle of system (2.19) is non-algebraic.
Example 2.8.
˙
x= x+ (x−y)(x2+2y2)2, y˙ =y−(x+y)(x2+2y2)2. We have
f(ϑ) =g(ϑ) =−1
4(cos 2ϑ−3)2, F(ϑ) =e4ϑ.
By Theorem2.1, we conclude that system has a hyperbolic stable limit cycle.
The cycle is written as r=√
ρ, ρ=
ρ∗0−8 Z ϑ
0
dτ e4τ(cos 2τ−3)2
e4ϑ, ρ∗0 = 8e
8π
e8π−1 Z 2π
0
dτ
e4τ(cos 2τ−3)2.
Just as in the previous example, it can be shown that the cycle is non-algebraic due to Theorem2.2.
Using Example 2.6 and Theorem 2.2, we can construct a Darboux system of any (odd) degree with an algebraic limit cycle.
Example 2.9.
˙
x=−x+ (x3−x2y+xy2−y3)(x2+y2)2k, y˙ =−y+ (x3+x2y+xy2+y3)(x2+y2)2k. The system has the algebraic limit cycle H(x,y)≡1−x2−y2=0.
As we already observed, in the presence of a limit cycle, all the remaining trajectories of system (1.2) are non-algebraic curves. In a neighborhood of the cycle, these trajectories are spirals with infinitely many helices and intersect a straight line transversal to the cycle infinitely many times. In this case, the corresponding algebraic equation has infinitely many roots, which is impossible.
Example2.6 shows that the coexistence of algebraic and non-algebraic curves is possible for cubic Darboux systems. Example2.7demonstrates that there exist cubic Darboux systems having no algebraic curve. With account taken of Theorem 2.2, this property is possessed by all Darboux systems (1.2) withn6=3 for whichg(ϑ)6=0, ϑ∈ [0, 2π](see Example2.8).
It was proved in [1] that system (1.2) is Darboux integrable and a formula for its first integral was given.
System (1.2) has no polynomial integral (and, generally, no integral defined on the whole phase plane) since its singular point at the origin is a node.
Under certain conditions, system (1.2) has a rational integral. In this case, all its trajectories are algebraic curves.
Recall that if system (1.1) hasN=n(n+1) +2 algebraic invariants then it admits a rational first integral (see [3]).
Example 2.10.
˙
x =x−x3, y˙= y−y3. (2.20)
The singular points of system (2.20) are:
O(0, 0),O1(1, 1),O2(−1, 1),O3(−1,−1),O4(1,−1)are star nodes;
O5(1, 0),O6(0, 1),O7(−1, 0),O8(0,−1)are hyperbolic saddles.
System (2.20) has 8 invariant straight lines x=0, x=±1, y=0, y=±1, y=±x, using which one can construct the rational Darboux integral
V(x,y) = y
2(1−x2) x2(1−y2). The phase portrait of system (2.20) is given in Figure2.1.
-4 -2 2 4x
-4 -2 2 4 y
Figure 2.1: The phase portrait of system (2.20)
In some cases, a rational integral for system (1.2) can be found with the use of a smaller number of invariants thanN.
Suppose that the homogeneous polynomials Pn(x,y),Qn(x,y) in system (1.2) satisfy the Cauchy–Riemann conditions: Pnx = Qny, Pny = −Qnx. Introducing the complex variable z= x+iy, we can write down this system in the form
˙
z=z+Pn(z), (2.21)
where Pn(z) is a complex polynomial of degree n : Pn(z) = Pn((z+z¯)/2,(z−z¯)/2i) + iQn((z+z¯)/2,(z−z¯)/2i).
Theorem 2.11 ([6]). Suppose that all the singular points of system (2.21) are star nodes and all the eigenvaluesλk1,2 = ωk are rationally commensurable: ωk ∈ ωQ, ω ∈ R, k = 1, 2, . . . ,n. Then system(2.21)admits a rational first integral.
In this case, the integral can be constructed with the use ofncomplex invariants of the form fk =z−zk, wherezk are the roots of the equationz+Pn(z) =0 (see [6]).
Example 2.12. Consider the system
˙
x =x−x5+10x3y2−5x4y, y˙= y−5x4y−10x2y3−y5, (2.22) which corresponds to the complex system ˙z= z−z5.
-1 1 x
-1 1 y
Figure 2.2: The phase portrait of the system (2.22)
The singular points of system (2.22) are found from the relation z−z5 = 0 and have the form O(0, 0), O1(1, 0), O2(0, 1), O3(−1, 0), O4(0,−1). The origin is an unstable star node; the remaining four singular points are stable star nodes. The system has 5 invariants
f1=z,f2,3=z±1,f4,5=z±i, using which we can construct the rational first integral V(x,y) = xy(x2−y2)
(x2+y2)4−(x2+y2)2+8x2y2. Figure2.2contains a phase portrait of system (2.22).
Acknowledgments
V. M. Cheresiz carried out research within the framework of the state contract of Sobolev In- stitute of Mathematics (project no. 0314-2019-0010). E. P. Volokitin carried out research within the framework of the state contract of Sobolev Institute of Mathematics (project no. 0314-2019- 0007).
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